Answer
For $x=-1$, $$(fog)'=-4\times2=-8$$
Work Step by Step
$f(u)=\Big(\frac{u-1}{u+1}\Big)^2$, $u=g(x)=\frac{1}{x^2}-1$, $x=-1$
Applying the Chain Rule: $$(fog)'=f'(g(x))g'(x)=f'(u)u'$$
We have
$$f'(u)=\Bigg[\Big(\frac{u-1}{u+1}\Big)^2\Bigg]'=2\Big(\frac{u-1}{u+1}\Big)\Big(\frac{u-1}{u+1}\Big)'$$ $$=2\Big(\frac{u-1}{u+1}\Big)\frac{(u-1)'(u+1)-(u-1)(u+1)'}{(u+1)^2}$$ $$=2\Big(\frac{u-1}{u+1}\Big)\frac{(u+1)-(u-1)}{(u+1)^2}=2\Big(\frac{u-1}{u+1}\Big)\frac{2}{(u+1)^2}$$ $$=\frac{4(u-1)}{(u+1)^3}$$
$$u'=\Big(\frac{1}{x^2}-1\Big)'=\frac{-1(x^2)'}{x^4}-0=-\frac{2x}{x^4}=-\frac{2}{x^3}$$
For $x=-1$, we have $$u'(-1)=-\frac{2}{(-1)^3}=-\Big(\frac{2}{-1}\Big)=2$$
Also, for $x=-1$, $u(-1)=\frac{1}{(-1)^2}-1=\frac{1}{1}-1=0$
Therefore, $$f'(0)=\frac{4(0-1)}{(0+1)^3}=\frac{-4}{1^3}=-4$$
That means,
$$(fog)'=-4\times2=-8$$
